MHT CET 2021 22th September Evening Shift | Dual Nature of Radiation Question 65 | Physics

The light of wavelength '$$\lambda$$' is incident on the surface of metal of work function $$\phi$$ and emits the electron. The maximum velocity of electron emitted is [$$\mathrm{m}=$$ mass of electron and $$\mathrm{h}=$$ Planck's constant, $$\mathrm{c}=$$ velocity of light]

$$\left[\frac{2(h c-\lambda)}{\mathrm{m} \lambda}\right]^{\frac{1}{2}}$$

$$\left[\frac{2(h c-\phi) \lambda}{m c}\right]$$

$$\left[\frac{2(h c-\lambda)}{m \lambda}\right]$$

$$\left[\frac{2(\mathrm{hc}-\lambda \phi)}{\mathrm{m} \lambda}\right]^{\frac{1}{2}}$$

MHT CET 2021 22th September Evening Shift

MCQ (Single Correct Answer)

-0

Photons of energy $$10 \mathrm{~eV}$$ are incident on a photosensitive surface of threshold frequency $$2 \times 10^{15} \mathrm{~Hz}$$. The kinetic energy in $$\mathrm{eV}$$ of the photoelectrons emitted is

[Planck's constant $$\mathrm{h}=6.63 \times 10^{34} \mathrm{~Js}$$ ]

8.29 eV

6.5 eV

4.2 eV

1.71 eV

MHT CET 2021 22th September Morning Shift

MCQ (Single Correct Answer)

-0

When a light of wavelength '$$\lambda$$' falls on the emitter of a photocells, maximum speed of emitted photoelectrons is '$$\mathrm{V}$$'. If the incident wavelength is changed to $$\frac{2 \lambda}{3}$$, maximum speed of emitted photoelectrons will be

less than $$\mathrm{V}(1.5)^{1 / 2}$$

$$\sqrt{\mathrm{V}}$$

grater than $$\mathrm{V}(1.5)^{\frac{1}{2}}$$

$$\mathrm{V}$$

MHT CET 2021 22th September Morning Shift

MCQ (Single Correct Answer)

-0

Kinetic energy of a proton is equal to energy '$$E$$' of a photon. Let '$$\lambda_1$$' be the de-Broglie wavelength of proton and '$$\lambda_2$$' is the wavelength of photon. If $$\frac{\lambda_1}{\lambda_2} \alpha E^n$$, then the value of '$$n$$' is

$$\frac{1}{2}$$

$$\frac{1}{4}$$

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